Question

A charge of 4.15 mC is placed at each corner of a square 0.310 m on a side.

1. Determine the magnitude of the force on each charge.

Answer 1

Since all corners have same charge therefore if we find force at one charge then all others charge will also experience the same force.
Now force on charge 4 are shown in the figure due to other charges.
Force on 4 due to 1
$F_{41} = \frac{kq_{1}*q_{4}}{a^{2}}$
On putting the values we get
$F_{41} = \frac{9*10^{9}*4.15*10^{-3}*4.15*10^{-3}}{0.31^{2}}$
F₄₁ = 1.613*10⁶ N
Now force on charge 4 due to charge 3
$F_{43} = \frac{kq_{3}*q_{4}}{a^{2}}$
$F_{43} = \frac{9*10^{9}*4.15*10^{-3}*4.15*10^{-3}}{0.31^{2}}$
F₄₃ = 1.613*10⁶ N
Now force due to charge 2
$F_{42} = \frac{kq_{2}*q_{4}}{(a\sqrt{2})^{2}}$
$F_{42} = \frac{9*10^{9}*4.15*10^{-3}*4.15*10^{-3}}{(0.31\sqrt{2})^{2}}$
F₄₂ = 8.06*10⁵ N
Now the force in X direction
F_X = F₄₁ + F₄₂Cos45 = 1.613*10⁶ + (8.06*10⁵)Cos45 = 21.83*10⁵ N
In the Y direction
F_Y = F₄₃ + F₄₂Sin45 = 1.613*10⁶ + (8.06*10⁵)Cos45 = 21.83*10⁵ N
Hence the magnitude of the force will be
$F = \sqrt{F_{X}^{2}+ F_{Y}^{2}}$
F = 30.87*10⁵ N
Similarly force on all charges will be 30.87*10⁵ N due to symmetry.